題目描述：

Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following problem as a punishment.

Find all integer solutions x (0 < x < 109) of the equation:

x = b·s(x)a + c, 

where a, b, c are some predetermined constant values and function s(x) determines the sum of all digits in the decimal representation of number x

The teacher gives this problem to Dima for each lesson. He changes only the parameters of the equation: a, b, c. Dima got sick of getting bad marks and he asks you to help him solve this challenging problem.

Input

The first line contains three space-separated integers: a, b, c (1 ≤ a ≤ 5; 1 ≤ b

Output

Print integer n — the number of the solutions that you've found. Next print n integers in the increasing order — the solutions of the given equation. Print only integer solutions that are larger than zero and strictly less than 109.

Examples

Input

3 2 8

Output

3
10 2008 13726 

Input

1 2 -18

Output

0

Input

2 2 -1

Output

4
1 31 337 967 

題目大意：公式中的s(x)是x的各個位數相加的和。如果由x來判斷等式是否成立，要計算的數太多，所以從s(x)這裡入手。

s（x）表示的是x的十進位制表示中所有的數字和，又因為x<10^9，所以x最多有9位，當每一位上都是9時s（x）有最大值81，所以s（x）的取值範圍為1到81。

要求輸出所有滿足條件的x的個數，並從小到大輸出x的值。

程式碼：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
const int maxn=500000;
int main()
{
   int a, b, c, s[10100];
   LL x, sx, num, sum;;
    int cnt=0;
   while(scanf("%d%d%d",&a,&b,&c)!=EOF)
   {
        for(int i=1;i<=81;i++)
        {
            sx=0;
            num=1;
            for(int j=1;j<=a;j++)
              num*=i;
            x=b*num+c;
            sum=x;
            while(x)
            {
              sx+=x%10;
              x/=10;
            }
            if(sx==i&&sum>0&&sum<1e9)//注意這裡sum的範圍一定要在判斷條件內，否則出錯
                s[cnt++]=sum;
        }
        cout<<cnt<<endl;
        if(cnt>0)
        {
        for(int i=0;i<cnt-1;i++)
            cout<<s[i]<<' ';
        cout<<s[cnt-1]<<endl;
        }
   }

   return 0;
}